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Na2CO3 + 2HCI → 2NaCI + H2O + CO2. Using the above equation, what volume...

Chemistry
JAMB 1981
Na2CO3 + 2HCI → 2NaCI + H2O + CO2. Using the above equation, what volume of carbondioxide, measured at s.t.p is liberated when 53g of sodium carbonate is dissolved in hydrochloric acid?
(1 mole of gas occupies 22.4 dm3 at s.t.p)
(Na = 23, C = 12, O = 16)
  • A. 44.8 dm3
  • B. 11.2 dm3
  • C. 100.1 dm3
  • D. 3.0 dm3
  • E. 22.4 dm3
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Correct Answer: Option B
Explanation
Na2CO3 + 2HCI → 2NaCI +H2O + CO2
53 gm Na2CO3 will therefore liberate (22.4)/(106) x (53)/(1) = 11.2 dm3
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